Differentiating Logarithmic Functions with Base e

Differential Calculus #8: Logarithmic Differentiation

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 2 videos by Darshan Senthil

Key Questions

$\frac{1}{x}$

Explanation:

${\log}_{e} \left(x\right)$ is commonly denoted as $\ln \left(x\right)$, the natural log.

$\implies \frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x}$

If you would like a proof, we can derive it from the limit definition:

${\lim}_{\delta x \to 0} \frac{f \left(x + \delta x\right) - f \left(x\right)}{\delta x}$

$= {\lim}_{\delta x \to 0} \frac{\ln \left(x + \delta x\right) - \ln \left(x\right)}{\delta x}$

$= {\lim}_{\delta x \to 0} \frac{\ln \left(\frac{x + \delta x}{x}\right)}{\delta x}$

$= {\lim}_{\delta x \to 0} \frac{1}{\delta x} \ln \left(1 + \frac{\delta x}{x}\right)$

$= {\lim}_{\delta x \to 0} \ln \left({\left(1 + \frac{\delta x}{x}\right)}^{\frac{1}{\delta x}}\right)$

$= {\lim}_{\delta x \to 0} \ln \left({\left(1 + \frac{\delta x}{x}\right)}^{\frac{1}{\delta x}}\right)$

$\text{Let } \tau \equiv \frac{\delta x}{x}$:

$= {\lim}_{\delta \tau \to 0} \ln \left({\left(1 + \tau\right)}^{\frac{1}{x \tau}}\right)$

$= {\lim}_{\delta \tau \to 0} \ln \left[{\left({\left(1 + \tau\right)}^{\frac{1}{\tau}}\right)}^{\frac{1}{x}}\right]$

$= \ln \left[{\left(e\right)}^{\frac{1}{x}}\right]$

$= \frac{1}{x} \ln \left(e\right)$

$= \frac{1}{x} \left(1\right)$

$= \frac{1}{x}$

• $y ' = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{{x}^{2}} = \frac{1 - \ln x}{{x}^{2}}$

This problem can also be solved by the Product Rule

$y ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g \left(x\right)$

The original function can also be rewritten using negative exponents.

$f \left(x\right) = \ln \frac{x}{x} = \ln \left(x\right) \cdot {x}^{-} 1$

$f ' \left(x\right) = \frac{1}{x} \cdot {x}^{-} 1 + \ln \left(x\right) \cdot - 1 {x}^{-} 2$

$f ' \left(x\right) = \frac{1}{x} \cdot \frac{1}{x} + \ln \left(x\right) \cdot - \frac{1}{x} ^ 2$

$f ' \left(x\right) = \frac{1}{x} ^ 2 - \ln \frac{x}{x} ^ 2$

$f ' \left(x\right) = \frac{1 - \ln \left(x\right)}{x} ^ 2$

• The answer would be $f ' \left(x\right) = \frac{1}{g} \left(x\right) \cdot g ' \left(x\right)$ or it can be written as $f ' \left(x\right) = \frac{g ' \left(x\right)}{g} \left(x\right)$.

To solve this derivative you will need to follow the chain rule which states:

$F \left(x\right) = f \left(g \left(x\right)\right)$ then $F ' \left(x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

Or without the equation, it the derivative of the outside(without changing the inside), times the derivative of the outside.

The derivative of $h \left(x\right) = \ln \left(x\right)$ is $h ' \left(x\right) = \frac{1}{x}$.

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